/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 The acceleration of a particle i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The acceleration of a particle increasing linearly with time \(t\) is \(b t\). The particle starts from the origin with an initial velocity \(v_{0}\). The distance travelled by the particle in time \(t\) will be (a) \(v_{0} t+\frac{1}{6} b t^{3}\) (b) \(v_{0} t+\frac{1}{6} b t^{2}\) (c) \(v_{0} t+\frac{1}{3} b t^{3}\) (d) \(v_{0} t+\frac{1}{3} b t^{2}\)

Short Answer

Expert verified
The correct option is (a) \( v_{0} t+\frac{1}{6} b t^{3} \).

Step by step solution

01

Identify the Given Data

We are given that the acceleration of the particle is\( a(t) = b \cdot t\). The particle starts from the origin with an initial velocity\( v_0\). We are asked to find the distance travelled by the particle in time \( t \).
02

Understand the Relation Between Acceleration and Velocity

To find the velocity, integrate the acceleration function. The acceleration \( a(t) = b \cdot t \) is a function of time, so the velocity \( v(t) \) can be found using the integral of acceleration:\[ v(t) = \int a(t) \, dt = \int b \cdot t \, dt = \frac{1}{2} b t^2 + C\]The constant \( C \) is the initial velocity \( v_0 \). So,\[ v(t) = \frac{1}{2} b t^2 + v_0\]
03

Calculate the Distance using Velocity

Distance or displacement is obtained by integrating the velocity function over time:\[ s(t) = \int v(t) \, dt = \int \left(\frac{1}{2} b t^2 + v_0\right) \, dt\]This leads to:\[ s(t) = \frac{1}{6} b t^3 + v_0 t + C'\]Since \( C' \) is the integration constant representing initial displacement, which is zero here (starting from the origin), we have:\[ s(t) = v_0 t + \frac{1}{6} b t^3\]
04

Verify the Solution with Given Options

Now, match the obtained expression for \( s(t) \), which is \( v_0 t + \frac{1}{6} b t^3 \), with the given options. The correct solution is option (a) \( v_{0} t+\frac{1}{6} b t^{3}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acceleration
Acceleration is a fundamental concept in kinematics that describes how the velocity of an object changes with time. In this particular exercise, the particle's acceleration changes linearly with time, meaning that the acceleration increases steadily as time progresses. This is represented by the function \( a(t) = b \cdot t \), where \( b \) is a constant indicating how quickly the acceleration increases. Acceleration is crucial because it influences both the velocity and the displacement of the particle over time. By understanding how acceleration works, we can predict the particle's motion and determine its velocity at any point in time. When dealing with kinematics, remember that acceleration isn't fixed—it can vary, as in this example, which directly affects overall movement.
Integral Calculus
Integral calculus is a key mathematical tool used to derive other quantities from given rates of change. In this scenario, we use integration to move from acceleration to velocity and then from velocity to displacement. To find velocity from acceleration, we integrate the acceleration function \( a(t) = b \cdot t \), leading to the velocity function \( v(t) = \frac{1}{2} b t^2 + v_0 \). This process essentially accumulates the small changes in acceleration over time to find how speed evolves. Similarly, integrating the velocity function yields the displacement: \( s(t) = v_0 t + \frac{1}{6} b t^3 \). By applying integral calculus, we transition from knowing the rate of change (acceleration) to understanding the total change in position over time (displacement), making it an invaluable process in solving kinematics problems.
Displacement
Displacement refers to the change in position of a particle over a given period. In kinematics, it's important to discern displacement from distance, as displacement considers direction, while distance does not. In this exercise, displacement is determined by integrating the velocity function, resulting in \( s(t) = v_0 t + \frac{1}{6} b t^3 \). This function tells us how far the particle has traveled from its starting point in time \( t \). Note that displacement is influenced by both the initial velocity \( v_0 \) and the time-dependent acceleration. Displacement helps convey not just how far a particle has moved but also in which direction, providing a complete picture of its motion from an initial position to its final location.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A body of mass \(m\) is accelerated uniformly from rest to a speed \(v\) in a time \(T\). The instantaneous power delivered to the body as a function of time is given by [AIEEE 2008] (a) \(\frac{1}{2} \frac{m v^{2}}{T^{2}} t^{2}\) (b) \(\frac{1}{2} \frac{m v^{2}}{T^{2}} t\) (c) \(\frac{m v^{2}}{\tau^{2}} t^{2}\) (d) \(\frac{m v^{2}}{T^{2}} t\)

The driver of a car moving with a speed of \(10 \mathrm{~ms}^{-1}\) sees a red light ahead, applies brakes and stops after covering \(10 \mathrm{~m}\) distance. If the same car were moving with a speed of \(20 \mathrm{~ms}^{-1}\), the same driver would have stopped the car after covering \(30 \mathrm{~m}\) distance. Within what distance the car can be stopped if travelling with a velocity of \(15 \mathrm{~ms}^{-1}\) ? Assume the same reaction time and the same deceleration in each case. (a) \(18.75 \mathrm{~m}\) (b) \(20.75 \mathrm{~m}\) (c) \(22.75 \mathrm{~m}\) (d) \(25 \mathrm{~m}\)

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement \(x\) is proportional to [UP SEE 2007] (a) \(x^{2}\) (b) \(\mathrm{e}^{x}\) (c) \(x\) (d) \(\log e^{x}\)

A ball is dropped from a bridge at a height of \(176.4 \mathrm{~m}\) over a river. After \(2 \mathrm{~s}\), a second ball is thrown straight downwards. What should be the initial velocity of the second ball so that both hit the water simultaneously? (a) \(2.45 \mathrm{~ms}^{-1}\) (b) \(49 \mathrm{~ms}^{-1}\) (c) \(14.5 \mathrm{~ms}^{-1}\) (d) \(24.5 \mathrm{~ms}^{-1}\)

A passenger arriving in a new town wishes to go from the station to a hotel located \(10 \mathrm{~km}\) away on a straight road from the station. A dishonest cabman takes him along a circuitoius path \(23 \mathrm{~km}\) long and reaches the hotel in \(28 \mathrm{~min}\). What are the average speed of the taxi and the magnitude of average velocity respectively (in \(\mathrm{km} / \mathrm{h}) ?\) (a) \(49.3\) and \(21.43\) (b) \(48.3\) and \(24.43\) (c) 21 and 20 (d) \(21.3\) and \(49.3\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.